Explanation

(d)
Given, $\tan^{- 1}{(x - 1) + \tan^{- 1}{x = \tan^{- 1}{3x - \tan^{- 1}{(x + 1)}}}}$
$\Rightarrow \ \tan^{- 1}{\left\lbrack \frac{(x - 1) + x}{1 - (x - 1)x} \right\rbrack = \tan^{- 1}\left\lbrack \frac{3x - (x + 1)}{1 + 3x(x + 1)} \right\rbrack}$
$\Rightarrow \ \ \ \left( 1 + 3x^{2} + 3x \right)(2x - 1) = \left( 1 - x^{2} + x \right)(2x - 1)$
$\Rightarrow \ \ (2x - 1)\left( 4x^{2} + 2x \right) = 0$
$\Rightarrow \ \ \ x = 0,\ \pm \frac{1}{2}$